Convolutions with differing arguments

Question

I kind of want to clear this up once and for all:

$g (t) * f(t) = \int g(u)f(t-u)du$

$g(at) * f(bt) = \int g(au)f(bt-u)du$ ???

Answer 1

The convolution $g*f$ is a single function, by definition: $(g*f)(t) = \int g(u)f(t-u)\,du$. Therefore $g(t)*f(t)$ doesn't really make sense. In particular, you can't plug in two different values at the same time to a function of one variable; so $g(at)*f(bt)$ definitely doesn't make sense. If you want to define new functions $G(t)=g(at)$ and $F(t)=f(bt)$ and consider $(G*F)(t) = \int G(u) F(t-u)\,du = \int g(au) f(b(t-u))\,du$, that's a reasonable thing to do.